Question

The particular integral of differential equation $f\left( D \right)y = {e^{ax}}$ where $f\left( D \right) = \left( {D - a} \right)g\left( D \right),g\left( a \right) \ne 0$ is:

$$A.{\text{ }}m{e^{ax}} \\\ B.{\text{ }}\dfrac{{{e^{ax}}}}{{g\left( a \right)}} \\\ C.{\text{ }}g\left( a \right){e^{ax}} \\\ D.{\text{ }}\dfrac{{x{e^{ax}}}}{{g\left( a \right)}} \\\$$

Answer 1

Hint: In order to solve the problem, first differentiate the given function partially or part by part. Further we will check the relation between $f\left( a \right)$ and $f'\left( a \right)$ and on the basis of their values we will use the formula of particular integral by substituting the constant term in the same function, by replacing d with constant a.

Complete step-by-step answer:
Given equation is $f\left( D \right)y = {e^{ax}}$ ---- (1)
Where $f\left( D \right) = \left( {D - a} \right)g\left( D \right),g\left( a \right) \ne 0$ ---- (2)
First let us find out $f\left( a \right)$ by using equation (2) substituting “a” in place of D.
$\because f\left( D \right) = \left( {D - a} \right)g\left( D \right) \\\ \Rightarrow f\left( a \right) = \left( {a - a} \right)g\left( D \right) \\\ \Rightarrow f\left( a \right) = \left( 0 \right)g\left( D \right) \\\ \Rightarrow f\left( a \right) = 0 \\\$
Now let us first differentiate equation (2) in order to find $f'\left( a \right)$
$\Rightarrow f'\left( D \right) = g\left( D \right) + \left( {D - a} \right)g'\left( D \right)$ --- (3)
First let us find out $f'\left( a \right)$ by using equation (3) substituting “a” in place of D
$\because f'\left( D \right) = g\left( D \right) + \left( {D - a} \right)g'\left( D \right) \\\ \Rightarrow f'\left( a \right) = g\left( a \right) + \left( {a - a} \right)g'\left( a \right) \\\ \Rightarrow f'\left( a \right) = g\left( a \right) + \left( 0 \right)g'\left( a \right) \\\ \Rightarrow f'\left( a \right) = g\left( a \right) + 0 \\\ \Rightarrow f'\left( a \right) = g\left( a \right) \\\$
From the above two result we have seen that $f\left( a \right) = 0\& f'\left( a \right) \ne 0$
So for particular integral we first differentiate equation (1) with respect to “a” and then find the ratio with $f'\left( a \right)$
Differentiating equation (1) we get:
$= \dfrac{d}{{da}}\left[ {{e^{ax}}} \right] \\\ = x{e^{ax}}.........{\text{(4) }}\left[ {\because \dfrac{d}{{dt}}\left[ {{e^{\alpha t}}} \right] = \alpha {e^{at}}} \right] \\\$
Also
$f'\left( a \right) = g\left( a \right)$
So, let us use the formula for a particular integral.
The particular integral is:
$= \dfrac{{x{e^{ax}}}}{{f'\left( a \right)}} \\\ = \dfrac{{x{e^{ax}}}}{{g\left( a \right)}} \\\$
Hence, the particular integral is $\dfrac{{x{e^{ax}}}}{{g\left( a \right)}}$
So, option D is the correct option.

Note- A general solution of an nth-order equation is a solution containing n arbitrary independent constants of integration. A particular solution is derived from the general solution by setting the constants to particular values, often chosen to fulfill set 'initial conditions or boundary conditions'. Particular integral is a part of the solution of the differential equation.